What would the image size of the moon be if you photographed it at Cassegrain focus of a 24-inch telescope with a focal length of 9600mm?
To determine the image size of the moon when photographed at the Cassegrain focus of a telescope, you need to use the formula for angular magnification. The formula is given as:
Angular Magnification (M) = Focal Length of Telescope (FT) / Focal Length of Eyepiece (FE)
In this case, the focal length of the telescope is 9600mm, but we don't have information about the focal length of the eyepiece. However, we can calculate the angular magnification if we assume a typical value for the focal length of the eyepiece.
A common choice for the focal length of an eyepiece is 10mm. With this assumption, we can calculate the angular magnification as follows:
Angular Magnification (M) = 9600mm / 10mm = 960
The angular magnification tells us how much larger an object appears when viewed through the telescope compared to the naked eye.
Now, to find the image size of the moon, we need to know the actual angular size of the moon. On average, the moon has an angular diameter of about 0.5 degrees or 30 arcminutes.
Image Size of the Moon = Angular Magnification * Angular Size of the Moon
Image Size of the Moon = 960 * 0.5 degrees = 480 degrees
Therefore, the image size of the moon would be 480 degrees when photographed at the Cassegrain focus of a 24-inch telescope with a focal length of 9600mm.